# Taylor餘項定理基礎 · 脈衝 R03

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations · 2026-10-15 · 100 題

欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。

[欄序與來源綁定資料](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[2,3],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0001/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 2. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[3,5,7],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0002/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 3. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[4,7,10,13],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0003/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 4. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[5,9,13,17,4],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0004/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 5. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[6,11,16,4,9,14],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0005/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 6. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[7,13],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0006/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 7. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[8,15,5],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0007/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 8. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[9,17,8,16],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0008/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 9. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[10,2,11,3,12],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0009/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 10. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[11,4,14,7,17,10],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0010/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 11. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[12,6],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0011/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 12. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[13,8,3],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0012/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 13. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[14,10,6,2],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0013/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 14. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[15,12,9,6,3],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0014/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 15. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[16,14,12,10,8,6],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0015/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 16. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[17,16],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0016/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 17. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[1,1,1],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0017/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 18. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[2,3,4,5],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0018/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 19. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[3,5,7,9,11],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0019/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 20. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[4,7,10,13,16,2],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0020/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 21. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[5,9],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0021/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 22. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[6,11,16],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0022/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 23. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[7,13,2,8],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0023/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 24. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[8,15,5,12,2],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0024/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 25. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[9,17,8,16,7,15],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。

查看答案與解釋([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Wave6第十層核對=([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0025/)

Taylor餘項定理 · qf-mod47-taylor-qualification · candidate





## 26. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[2,3],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(1, 1, [2, 3/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(1, 1, [2, 3/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0026/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 27. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[3,5,7],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(2, 2, [3, 5/2, 7/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(2, 2, [3, 5/2, 7/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0027/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 28. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[4,7,10,13],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(3, 3, [4, 7/2, 5/2, 13/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(3, 3, [4, 7/2, 5/2, 13/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0028/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 29. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[5,9,13,17,4],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0029/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 30. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[6,11,16,4,9,14],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0030/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 31. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[7,13],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(1, 1, [7, 13/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(1, 1, [7, 13/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0031/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 32. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[8,15,5],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(2, 2, [8, 15/2, 5/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(2, 2, [8, 15/2, 5/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0032/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 33. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[9,17,8,16],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(3, 3, [9, 17/2, 2, 2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(3, 3, [9, 17/2, 2, 2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0033/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 34. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[10,2,11,3,12],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(4, 4, [10, 1, 11/4, 3/8, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(4, 4, [10, 1, 11/4, 3/8, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0034/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 35. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[11,4,14,7,17,10],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0035/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 36. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[12,6],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(1, 1, [12, 3], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(1, 1, [12, 3], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0036/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 37. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[13,8,3],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(2, 2, [13, 4, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(2, 2, [13, 4, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0037/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 38. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[14,10,6,2],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(3, 3, [14, 5, 3/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(3, 3, [14, 5, 3/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0038/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 39. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[15,12,9,6,3],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(4, 4, [15, 6, 9/4, 3/4, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(4, 4, [15, 6, 9/4, 3/4, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0039/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 40. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[16,14,12,10,8,6],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0040/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 41. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[17,16],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(1, 1, [17, 8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(1, 1, [17, 8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0041/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 42. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[1,1,1],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(2, 2, [1, 1/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(2, 2, [1, 1/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0042/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 43. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[2,3,4,5],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(3, 3, [2, 3/2, 1, 5/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(3, 3, [2, 3/2, 1, 5/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0043/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 44. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[3,5,7,9,11],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0044/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 45. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[4,7,10,13,16,2],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0045/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 46. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[5,9],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(1, 1, [5, 9/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(1, 1, [5, 9/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0046/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 47. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[6,11,16],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(2, 2, [6, 11/2, 4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(2, 2, [6, 11/2, 4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0047/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 48. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[7,13,2,8],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(3, 3, [7, 13/2, 1/2, 1], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(3, 3, [7, 13/2, 1/2, 1], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0048/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 49. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[8,15,5,12,2],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0049/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 50. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[9,17,8,16,7,15],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。

查看答案與解釋(5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)

Wave6第十層核對=(5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0050/)

Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate





## 51. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[2,3,2],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(2, 1, 4, 1/4, 是, 1/2, 1/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(2, 1, 4, 1/4, 是, 1/2, 1/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0051/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 52. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[3,5,7,3],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(3, 2, 18, 1/4, 是, 3/8, 3/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(3, 2, 18, 1/4, 是, 3/8, 3/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0052/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 53. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[4,7,10,13,4],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(4, 3, 96, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(4, 3, 96, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0053/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 54. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[5,9,13,17,4,5],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(5, 4, 600, 1/4, 是, 5/32, 5/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(5, 4, 600, 1/4, 是, 5/32, 5/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0054/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 55. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[6,11,16,4,9,14,6],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(6, 5, 4320, 1/4, 是, 3/32, 3/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(6, 5, 4320, 1/4, 是, 3/32, 3/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0055/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 56. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[7,13,7],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(2, 1, 14, 1/4, 是, 7/4, 7/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(2, 1, 14, 1/4, 是, 7/4, 7/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0056/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 57. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[8,15,5,8],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(3, 2, 48, 1/4, 是, 1, 1, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(3, 2, 48, 1/4, 是, 1, 1, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0057/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 58. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[9,17,8,16,9],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(4, 3, 216, 1/4, 是, 9/16, 9/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(4, 3, 216, 1/4, 是, 9/16, 9/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0058/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 59. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[10,2,11,3,12,10],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(5, 4, 1200, 1/4, 是, 5/16, 5/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(5, 4, 1200, 1/4, 是, 5/16, 5/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0059/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 60. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[11,4,14,7,17,10,11],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(6, 5, 7920, 1/4, 是, 11/64, 11/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(6, 5, 7920, 1/4, 是, 11/64, 11/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0060/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 61. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[12,6,12],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(2, 1, 24, 1/4, 是, 3, 3, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(2, 1, 24, 1/4, 是, 3, 3, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0061/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 62. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[13,8,3,13],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(3, 2, 78, 1/4, 是, 13/8, 13/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(3, 2, 78, 1/4, 是, 13/8, 13/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0062/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 63. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[14,10,6,2,14],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(4, 3, 336, 1/4, 是, 7/8, 7/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(4, 3, 336, 1/4, 是, 7/8, 7/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0063/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 64. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[15,12,9,6,3,15],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(5, 4, 1800, 1/4, 是, 15/32, 15/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(5, 4, 1800, 1/4, 是, 15/32, 15/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0064/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 65. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[16,14,12,10,8,6,16],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(6, 5, 11520, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(6, 5, 11520, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0065/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 66. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[17,16,17],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(2, 1, 34, 1/4, 是, 17/4, 17/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(2, 1, 34, 1/4, 是, 17/4, 17/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0066/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 67. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[1,1,1,18],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(3, 2, 108, 1/4, 是, 9/4, 9/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(3, 2, 108, 1/4, 是, 9/4, 9/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0067/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 68. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[2,3,4,5,19],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(4, 3, 456, 1/4, 是, 19/16, 19/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(4, 3, 456, 1/4, 是, 19/16, 19/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0068/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 69. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[3,5,7,9,11,20],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(5, 4, 2400, 1/4, 是, 5/8, 5/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(5, 4, 2400, 1/4, 是, 5/8, 5/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0069/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 70. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[4,7,10,13,16,2,21],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(6, 5, 15120, 1/4, 是, 21/64, 21/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(6, 5, 15120, 1/4, 是, 21/64, 21/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0070/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 71. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[5,9,22],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(2, 1, 44, 1/4, 是, 11/2, 11/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(2, 1, 44, 1/4, 是, 11/2, 11/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0071/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 72. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[6,11,16,23],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(3, 2, 138, 1/4, 是, 23/8, 23/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(3, 2, 138, 1/4, 是, 23/8, 23/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0072/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 73. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[7,13,2,8,24],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(4, 3, 576, 1/4, 是, 3/2, 3/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(4, 3, 576, 1/4, 是, 3/2, 3/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0073/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 74. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[8,15,5,12,2,25],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(5, 4, 3000, 1/4, 是, 25/32, 25/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(5, 4, 3000, 1/4, 是, 25/32, 25/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0074/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 75. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[9,17,8,16,7,15,26],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。

查看答案與解釋(6, 5, 18720, 1/4, 是, 13/32, 13/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Wave6第十層核對=(6, 5, 18720, 1/4, 是, 13/32, 13/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0075/)

Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate





## 76. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[2,3,2],"center":0,"x":[1,2],"order":1,"M":4}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 1/2, 4, 4, 1/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 1/2, 4, 4, 1/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0076/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 77. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[3,5,7,3],"center":0,"x":[1,2],"order":2,"M":17}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 3/8, 18, 17, 17/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 3/8, 18, 17, 17/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0077/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 78. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[4,7,10,13,4],"center":0,"x":[1,2],"order":3,"M":96}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 1/4, 96, 96, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 1/4, 96, 96, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0078/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 79. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[5,9,13,17,4,5],"center":0,"x":[1,2],"order":4,"M":599}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 5/32, 600, 599, 599/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 5/32, 600, 599, 599/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0079/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 80. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[6,11,16,4,9,14,6],"center":0,"x":[1,2],"order":5,"M":4320}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 3/32, 4320, 4320, 3/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 3/32, 4320, 4320, 3/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0080/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 81. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[7,13,7],"center":0,"x":[1,2],"order":1,"M":13}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 7/4, 14, 13, 13/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 7/4, 14, 13, 13/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0081/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 82. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[8,15,5,8],"center":0,"x":[1,2],"order":2,"M":48}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 1, 48, 48, 1, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 1, 48, 48, 1, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0082/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 83. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[9,17,8,16,9],"center":0,"x":[1,2],"order":3,"M":215}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 9/16, 216, 215, 215/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 9/16, 216, 215, 215/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0083/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 84. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[10,2,11,3,12,10],"center":0,"x":[1,2],"order":4,"M":1200}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 5/16, 1200, 1200, 5/16, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 5/16, 1200, 1200, 5/16, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0084/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 85. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[11,4,14,7,17,10,11],"center":0,"x":[1,2],"order":5,"M":7919}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 11/64, 7920, 7919, 7919/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 11/64, 7920, 7919, 7919/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0085/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 86. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[12,6,12],"center":0,"x":[1,2],"order":1,"M":24}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 3, 24, 24, 3, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 3, 24, 24, 3, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0086/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 87. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[13,8,3,13],"center":0,"x":[1,2],"order":2,"M":77}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 13/8, 78, 77, 77/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 13/8, 78, 77, 77/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0087/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 88. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[14,10,6,2,14],"center":0,"x":[1,2],"order":3,"M":336}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 7/8, 336, 336, 7/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 7/8, 336, 336, 7/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0088/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 89. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[15,12,9,6,3,15],"center":0,"x":[1,2],"order":4,"M":1799}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 15/32, 1800, 1799, 1799/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 15/32, 1800, 1799, 1799/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0089/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 90. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[16,14,12,10,8,6,16],"center":0,"x":[1,2],"order":5,"M":11520}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 1/4, 11520, 11520, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 1/4, 11520, 11520, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0090/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 91. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[17,16,17],"center":0,"x":[1,2],"order":1,"M":33}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 17/4, 34, 33, 33/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 17/4, 34, 33, 33/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0091/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 92. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[1,1,1,18],"center":0,"x":[1,2],"order":2,"M":108}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 9/4, 108, 108, 9/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 9/4, 108, 108, 9/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0092/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 93. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[2,3,4,5,19],"center":0,"x":[1,2],"order":3,"M":455}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 19/16, 456, 455, 455/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 19/16, 456, 455, 455/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0093/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 94. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[3,5,7,9,11,20],"center":0,"x":[1,2],"order":4,"M":2400}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 5/8, 2400, 2400, 5/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 5/8, 2400, 2400, 5/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0094/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 95. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[4,7,10,13,16,2,21],"center":0,"x":[1,2],"order":5,"M":15119}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 21/64, 15120, 15119, 15119/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 21/64, 15120, 15119, 15119/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0095/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 96. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[5,9,22],"center":0,"x":[1,2],"order":1,"M":44}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 11/2, 44, 44, 11/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 11/2, 44, 44, 11/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0096/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 97. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[6,11,16,23],"center":0,"x":[1,2],"order":2,"M":137}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 23/8, 138, 137, 137/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 23/8, 138, 137, 137/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0097/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 98. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[7,13,2,8,24],"center":0,"x":[1,2],"order":3,"M":576}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 3/2, 576, 576, 3/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 3/2, 576, 576, 3/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0098/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 99. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[8,15,5,12,2,25],"center":0,"x":[1,2],"order":4,"M":2999}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(undersized_M, 25/32, 3000, 2999, 2999/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(undersized_M, 25/32, 3000, 2999, 2999/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0099/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate





## 100. Taylor餘項定理

center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[9,17,8,16,7,15,26],"center":0,"x":[1,2],"order":5,"M":18720}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。

actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。

查看答案與解釋(valid_bound, 13/32, 18720, 18720, 13/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

Wave6第十層核對=(valid_bound, 13/32, 18720, 18720, 13/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0100/)

Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations/validation-report.json)